How do you rationalize #1/(1+sqrt3-sqrt5)#?

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Answer 1 · 2015-05-24T17:26:13

The quick answer is: multiply both numerator (top) and denominator by:

#(1+sqrt(3)+sqrt(5))(1-sqrt(3)-sqrt(5))(1-sqrt(3)+sqrt(5))#

Quick to say, but a little slow to do...

Let's take it one step at a time...

#(1+sqrt(3)-sqrt(5))(1+sqrt(3)+sqrt(5))#

#=(1+sqrt(3))^2-sqrt(5)^2#

#=1+2sqrt(3)+3-5#

#=-1+2sqrt(3)#

For slightly complex reasons, we can reverse the sign on all occurences of #sqrt(3)# to deduce:

#(1-sqrt(3)-sqrt(5))(1-sqrt(3)+sqrt(5))=-1-2sqrt(3)#

So

#(1+sqrt(3)-sqrt(5))(1+sqrt(3)+sqrt(5))(1-sqrt(3)-sqrt(5))(1-sqrt(3)+sqrt(5))#

#=(-1+2sqrt(3))(-1-2sqrt(3))#

#=(1-2sqrt(3))(1+2sqrt(3))#

#=1^2-(2sqrt(3))^2#

#=1-12=-11#

So

#1/(1+sqrt(3)-sqrt(5)) = -((1+sqrt(3)+sqrt(5))(1-sqrt(3)-sqrt(5)) (1-sqrt(3)+sqrt(5)))/11#